This seems like a great solution, but I'll need to take some time to properly "digest" it (I've never seen this technique so far, and I want to make sure I properly understand what happens)
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Gabi PurcaruJul 18 '12 at 8:58

we have, integrating this equation on $[0,1]$, $$\frac{1}{n-1}-C_{n-2}=C_n.$$ Hence we have a recurrence relation for the $C_n$'s. Let's see what this gives. We have
$$C_0=\int_0^1 \frac{ \mbox{d}x}{1+x^2} = \arctan(1) = \frac{\pi}{4},$$
and

$$C_1=\int_0^1 \frac{x\ \mbox{d}x}{1+x^2} = \frac{1}{2}\log2.$$

Now using the recurrence we find

$C_0=\frac{\pi}{4}$

$C_1=\frac{1}{2}\log2$

$C_2=1-\frac{\pi}{4}$

$C_3=\frac{1}{2}-\frac{1}{2}\log2$

$C_4 = -1+\frac{1}{3} +\frac{\pi}{4}$

$C_5=-\frac{1}{2}+\frac {1}{4} +\frac{1}{2}\log 2$

$C_6=1-\frac{1}{3}+\frac{1}{5} - \frac{\pi}{4}$

...

Now if we define

$$A_n = \sum_{k=1}^n\frac{(-1)^{k-1}}{2k},$$
$$B_n = \sum_{k=1}^n\frac{(-1)^{k-1}}{2k-1},$$
it is easy to see by induction that $$C_{2n} = (-1)^n\left(\frac{\pi}{4}-B_n\right)$$
and

$$C_{2n-1} = (-1)^{n-1}\left(\frac{\log 2}{2} -A_{n-1}\right).$$

Notice that $A_n \rightarrow \frac{1}{2}\log2$ as $n\rightarrow \infty$ [recall the series expansion of $\log(1+x)$], and that $B_n \rightarrow \frac{\pi}{4}$ as $n\to \infty$ [recall the series expansion of $\arctan x$, which you can get by integrating $(1+x^2)^{-1}$].

Let us examine the partial sums

$$I_{2N}=\sum_{n=1}^{2N} \frac{(-1)^{n-1}}{n}C_n.$$

We can split this into the odd-labeled terms and the even-labeled terms, as

I edited your answer. If you are interesting in $TeX$ code you can see it clicking on edit button. For some basic information about writing math at this site see e.g. here, here, here and here. And nice answer! (+1)
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CortizolJun 29 '13 at 20:13